Question

On a coordinate plane, a line goes through (negative 3, negative 3) and (negative 1, 5). What is the equation of the line parallel to the given line with an x-intercept of 4? y = - x +

Answer 1

Answer:

y=4x-16

Explanation:

The general form of the equation of a line is: y=mx+x, where m is the gradient and c is the y-intercept.

Given points  (-3,-3) and (-1,5)

We are to find the equation of the line parallel to the given line with an x-intercept of 4.

First, we determine the value of the gradient of the line.

Gradient,

$m=\frac{y_2-y_1}{x_2-x_1} \\=\frac{5-(-3)}{-1-(-3)} \\=\frac{5+3}{-1+3}\\=\frac{8}{2}\\m=4$

Two lines are parallel if their gradients $m_1 \: and \: m_2$ are equal.

The gradient of the line parallel to the given line $m_2=4$

Since the x-intercept of the new line is 4, a point on the new line, $(x_1,y_1)=(4,0)$

Substituting $(x_1,y_1)=(4,0)$ into the equation of a line: , $y-y_1=m(x-x_1),$we have:

$y-0=4(x-4)\\ Therefore:\\y=4x-16$

Therefore, the equation of the line is: y=4x-16



Answer 2

Answer:

y = 4x-16

Explanation:

The coordinate points are: (-3,-3)(-1,5)

to get the slope of the line, we use the formula

$m= \frac{x_{2} - x_[1}{y_{2}-y_{1}}=\frac{2}{8}=4$

since the two lines are parallel, it means that they have equal slopes.

hence, the slope of the second line = 4

Since the x intercept of the second line is 4, it means the point on the second line is (4, 0)

using the formula

$y-y_{1}=m(x-x_{1})$

$y-0=4(x-4)$

y=4x-16